If we have an action of the form:

$$ S^{(0)} =\int d^Dx\sqrt{-G}\left(R+\alpha (\partial \phi)^2 + \frac{1}{6} e^{2 \alpha \phi}H_{mnp}H^{mnp}\right).$$

where $H_{mnp}=\partial_m B_{np}+\partial_n B_{pm}+\partial_p B_{mn}$ , $B_{mn}$ is an anti-symmetric tensor and $\alpha= \frac{4}{D-2}$ . This is the effective action of lowest order in the, so called, s parametrization in the section of massless particles in the theory of quantum strings. In order to find the equations for the anti-symetric tensor I did the following:

$$ \delta S^{(0)} =\int d^Dx\sqrt{-G}\delta\left(R+\alpha (\partial \phi)^2 + \frac{1}{6} e^{2 \alpha \phi}H_{mnp}H^{mnp}\right) $$

$$ \delta S^{(0)} =\int d^Dx\sqrt{-G}\left( \frac{1}{6} e^{2 \alpha \phi} \delta (H_{mnp}H^{mnp})\right) $$

$$ \delta S^{(0)} =\int d^Dx\sqrt{-G}\left( \frac{1}{3} e^{2 \alpha \phi} \delta (H_{mnp})H^{mnp}\right) $$


$$\delta (H_{mnp})H^{mnp} =\partial_m\delta(B_{np})[H^{mnp}+H^{pmn}+H^{npm}] $$

$$ \delta S^{(0)} =\int d^Dx\sqrt{-G}\left( \frac{1}{3} e^{2 \alpha \phi} \partial_m\delta(B_{np})[H^{mnp}+H^{pmn}+H^{npm}] \right) $$

From here I integrated by parts:

$$ \delta S^{(0)} =\int d^{D-1}x\sqrt{F}\left( \frac{1}{3} e^{2 \alpha \phi} \delta(B_{np}) [H^{mnp}+H^{pmn}+H^{npm}] \right) \Big|_a^b \ - \int d^Dx\sqrt{-G}\left( \frac{1}{3} e^{2 \alpha \phi} \delta(B_{np})[H^{mnp}+H^{pmn}+H^{npm}]_{,m} \right) $$

and the first integral has its limits of integration that I labelled as $a$ and $b$. Do I get rid of the first term as it´s being evaluated in the respective limits (something similar to what we do in the anti symmetric Faraday tensor) and the 2nd term yields me the equation Im looking for? Or is my procedure wrong?


It´s most likely wrong because the metric is D-dimentional but I dont know how to do it any other way.

Edit 2:

The equations I got for the metric field:

$$ -\frac{1}{2}G_{\mu\nu}[R+ \alpha (\partial \phi)^2+\frac{1}{6} e^{2 \alpha \phi} H_{mnp}H^{mnp}]+R_{\mu\nu}+\alpha \partial_\mu \phi \partial_\nu \phi + \frac{1}{2} e^{2 \alpha \phi} H_{\mu np}H_{\nu}^{\space np} = 0$$

And for the field $\phi$:

$$ - \frac{1}{3} e^{2 \alpha \phi} H_{mnp}H^{mnp} + 2 \alpha \square \phi = 0$$

Edit 3:

The equation I got for the anti-symmetric tensor:

$$ \partial_m [e^{2 \alpha \phi} (H^{mnp}+H^{pmn}+H^{npm})] = 0 $$

  • $\begingroup$ What's your point about the action being d dimensional?? $\endgroup$
    – Noone
    Nov 4, 2020 at 13:24
  • $\begingroup$ In the case of finding the equations of the metric field for example, I couldn´t do $R=R_{\mu\nu} G^{\mu\nu}$ right? Since we have a D dimensional metric it would have D indicies. I know its not the specific case I asked for but maybe it had some implications here. $\endgroup$ Nov 4, 2020 at 15:37
  • $\begingroup$ I think that there are some misconceptions or a language problem. What do you mean with "D indices"? $\endgroup$
    – Noone
    Nov 4, 2020 at 15:40
  • $\begingroup$ Also, the Ricci scalar is defined as the contraction of the Ricci tensor with the metric:$R = g^{μν}R_{μν}$. $\endgroup$
    – Noone
    Nov 4, 2020 at 15:41
  • $\begingroup$ No, forget about it, Im not going to write comments again when Im dead tired, For some odd reason my brain thought that with D dimentions the associated metric tensor needed D indicies instead of each index having D possibilities. Sorry my bad. $\endgroup$ Nov 4, 2020 at 17:07

2 Answers 2


The variation of the field $B$ vanishes at the boundary thus the total derivative term vanishes. Now regarding the last term, the derivative should also act on the exponential. When you apply Leibniz rule you have:

$$\partial(e^{2 \alpha \phi}\delta(B_{np}) [H^{mnp}+H^{pmn}+H^{npm}] )=0 $$

as a total divergence term. The derivative acts on three terms, you forgot the exponential.


$$\partial_m [e^{2 \alpha \phi} (H^{mnp}+H^{pmn}+H^{npm})] = 0$$

Shouldn't be?

$$\nabla_m [e^{2 \alpha \phi} (H^{mnp}+H^{pmn}+H^{npm})] = 0$$

  • 2
    $\begingroup$ For a torsion-free space you would be correct, but you must mention that. For spaces with torsion the Levi-Civita connections are not symmetric w.r.t. the covariant indicies, hence they dont cancel each other out. $\endgroup$ Jan 6, 2021 at 14:42

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